Solution

There are just two pieces of information that are important to solving this problem. That is, the knowledge of the definition of logarithms is being assumed.

However, just in case, logab is the power to which a is to be raised to give b:

alogab = b.

If logab = c, i.e., if ac = b, then a = b1/c which tells us that b to the power of 1/c gives a. In other words, logba = 1/c = 1/logab.

It follows that logab and logba are reciprocal numbers so that their product is exactly 1:

logab × logba = 1.

Also, from the definition of the logarithm it follows that logab = 1 only if a = b.

Now, let's return to the problem:

logπ3 + log3π > 2.

So, you should know two facts. First, whatever else can be said of the famous constant π, the only important thing that is relevant here is that π ≠ 3. This is important because it shows that logπ3 ≠ 1 ≠ log3π. The significance of this observation comes to light when we consider the second important fact, namely, a special form of the AM-GM (arithmetic mean - geometric mean) inequality: for every x > 0,

x + 1/x ≥ 2,

with the equality only if x = 1. In general, (a + b)/2 ≥ √ab, with equality only when a = b. Taking a = x and b = 1/x we obtain the aforementioned special case.

Of course, it could have been proved directly. Since x is assumed positive, the inequality obtained by multiplying by,dividing by,adding to both sides,multiplying by x, i.e., x² + 1 ≥ 2x is the same as x² - 2x + 1 = (x - 1)² ≥ 0, which is obvious.

Now, the solution to the problem becomes transparent. Since logπ3 and log3π are two reciprocals, their sum is never less than 2 and would only be equal to 2 in case each of them were 1, but, as we saw, they are not. The problem is solved.

Note: it is clear from the solution that appearance of π in the problem is a red herring. Replacing π with any positive number different from 3 would give the same result and in exactly the same manner.